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Claim: The fractional part of $\frac{2z}{i}$, where $z$ and $i$ are finite is no more than $\frac{i-2}{i}$ when $i$ is even and no more than $\frac{i-1}{i}$ when $i$ is odd. Here $z,i$ are non-negative integers. To make the problem non-trivial, assume $i \ge 3$.

Example: If $i=4$, then $2z$ can be $0,2,4,6,...12...$. The fractional parts for these numbers then are $0,2/4$. The largest of this is 2/4 which is $\frac{i-2}{i}$.

Second example: If $i=7$, then $2z$ can be $0,2,4,6,...42...$. The fractional parts for these numbers then are $0,2/7, 4/6,6/7,3/7,5/7$. The largest of this is 6/7 which is $\frac{i-1}{i}$.

I cannot write this simple result in the form of a rigorous proof. How should I?

UPDATE: I have provided an answer myself. Corrections are welcome.

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  • $\begingroup$ Why has this been downvoted without any explanation? $\endgroup$ – bissi Jan 24 '20 at 14:33
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If $b$ is even, $\{\frac{2a}{b} \}$ is one of $ \{ 0, \frac{2}{b}, \frac{4}{b}, \frac{6}{b}, \ldots, ... \frac{b-2}{b} \}$. The maximum of these is $\frac{b-2}{b}$. If $b$ is odd, $\{\frac{2a}{b} \}$ is one of $ \{ 0, \frac{2}{b}, \frac{4}{b}, \frac{6}{b}, \ldots, ... \frac{b-1}{b} \}$. The maximum of these is $\frac{b-1}{b}$.

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